This is one of the most counter-intuitive probability puzzles ever to be part of a game show format. It’s brought countless brainiacs undone, and left even the most highly educated specimens making the completely wrong decision, all the while vehemently insisting that they’re completely right. Why? What’s so hard about it? And why does the so-called Monty Hall Problem seem to slap our natural, instinctive human decision-making right in the face?

Good, if slightly melodramatic, question. Let’s explore it for a bit, and see if we can figure it out. But first, what exactly IS “The Monty Hall Problem”? Well, it goes something a little like this…

You’re on a game show. Congratulations – so far, so good. On the set are three closed, non-transparent doors – Door A, Door B and Door C. Behind one of them is a new car, an behind the other two doors are goats. Which you don’t want. Your host (who knows what’s behind each door) asks you to choose the one door that you think the car is behind. You choose Door B. The host opens one of the two remaining doors – say Door C – to reveal a goat. That now leaves two unopened doors – Door A and Door B, one of which does have the car behind it. You originally chose Door B. You’re now asked if you want to change your decision, and switch to Door A. What do you do?

If you’re like most people, your first reaction will be “Oh, I’ll stick with the door I originally chose. Now that there are only 2 doors left, I’ve got a 1-in-2 chance of being right  And besides, I’d hate to switch and then be wrong! No, I’m backing my original decision”.

Here’s the thing… at this point in the game, you do not have a 1-in-2 chance of being right – it’s not 1-in-2 at all – and statistically, you should always switch.

By the way, it’s called the Monty Hall problem because it came to fame on the show Let’s Make A Deal, which was hosted by Monty Hall. Although I think they rarely used actual goats.

The Monty Hall Problem, seen here in its native habitat – the ‘Let’s Make A Deal’ studio. That’s Monty on the right, doing the slightly awkward Shop Window Mannequin Hands.

So why don’t you have a 1-in-2 chance of being right, after that first door’s been opened and there are just two doors left? What chance is it, then? And why should you always switch?

Before we go any further, we need to be clear about the restrictions on the host’s behaviour:

1. The host must always open a door that was not picked by the contestant.
2. The host must always open a door to reveal a goat and never the car.
3. The host must always offer the chance to switch between the originally chosen door and the remaining closed door.

So, in the example above, at the start of the game, you chose Door B, which had a 1-in-3 chance of being the winning door. This means that between them (that is to say, combined) Doors A and C had a 2-in-3 chance of being the winning door.

But then, one of the non-winning doors was opened, and you think “Aha! Now the odds have changed, and they’ve changed in my favour! I had a 1-in-3 chance a moment ago, and now I have a 1-in-2 chance!”

And that’s where you’re wrong.

And this, I think, is the bit that’s hard to get our heads around… the host opening one of the two non-winning doors does not change the 1-in-3 odds of your original decision.

Again, the host opening one of the two non-winning doors does not change the 1-in-3 odds of your original decision. Your original decision to choose Door B was made when there were three options available, and now there are two. Your decision was not made when there were two options available, and therefore it is not a 1-in-2 chance. After the first door has been opened, you still have the same chance you had at the start; a 1-in-3 (or roughly 33%) chance of having the winning door. But that doesn’t quite feel right; it “makes more sense” to us to think that if there are 2 options left, then you must have a 1-in-2 chance. “It stands to reason!” Ah, but it doesn’t. And that’s why this puzzle is such a head scratcher.

And that’s why you don’t have a 1-in-2 chance of being right, after that first door’s been opened.

So, even though one door has been opened, and just one remains, you still have a 1-in-3 chance of winning. Logically, then, the one door that you haven’t chosen (but which remains in play) has a 2-in-3 (or roughly 66%) chance of being the winning door. So switching from your chosen door to the one remaining door you didn’t choose actually DOUBLES your odds of winning – you go from a 1-in-3 chance of winning (33%) to a 2-in-3 chance of winning (66%).

And THAT’s why you should always switch.

Derren Brown also devoted some time to this in his most excellent book Tricks of The Mind, which I’ll be reviewing here next week. (Spoiler alert: I liked it).

He outlines another scenario, which I think is helpful if you’re still thinking that the car being behind your originally chosen door is a 50/50 chance….

He asks you to imagine that the problem involves 100 doors, and not just 3. There is still a car behind one of these doors; the other 99 contain worthless prizes. You make your choice – say Door #56. At this point, when all doors are closed, the chance that Door #56 has the car behind it is 1-in-100; a 1% chance. According to the rules of the game, the host now opens 98 doors which the car is NOT behind, leaving you with two unopened doors, as per the original problem. The two remaining unopened doors are your originally chosen door (Door #56) and, say, Door #73. You’re now asked if you want to switch your decision from Door #56 to Door #73. What do you do?

When you had 100 doors to choose from, your choice had, by definition, a 1% chance of concealing the car. Now that the host – who knows where the car is, remember – has opened every door except yours and Door #73, the odds have changed drastically. The car is definitely behind one of the only two doors left unopened from the original lot of 100. At the start of the game, your chance of choosing the winning door was 1%. Now that there are only 2 doors left unopened, and the car is behind one of them, the chance that it’s behind Door #73 is 99%!

Viewing it in this light makes the whole thing seem a bit more intuitive or “logical”, doesn’t it?

If you’re still scratching your head about this conundrum, there’s a really interesting Wikipedia article here, another explanation along with a little program that lets you play the game here, and a nifty animation here that probably explains it in a far snappier way than I just have.

Anyway, that’s the Monty Hall Problem, or, as I’ve just decided to call it, Monty Hall’s Life of Brain. I hope I’ve managed to shed a little light on this confounding puzzle. Next week, as promised, a review of Derren Brown’s book, in which he explores logical problems, brain training, memory boosting techniques, neuro-linguistic programming, hypnosis… and clairvoyance! Until then…